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The authors of the paper "Driving Performance While Using a Mobile Phone: A Simulation Study of Greek Professional Drivers" (Transportation Research Part \(F\) [2016]: 164-170) describe a study to evaluate the effect of mobile phone use by taxi drivers in Greece. Fifty taxi drivers drove in a driving simulator where they were following a lead car. The drivers were asked to carry on a conversation on a mobile phone while driving, and the following distance (the distance between the taxi and the lead car) was recorded. The sample mean following distance was 3.20 meters and the sample standard deviation was 1.11 meters. a. Construct and interpret a \(95 \%\) confidence interval for \(\mu,\) the population mean following distance while talking on a mobile phone for the population of taxi drivers. b. What assumption must be made in order to generalize this confidence interval to the population of all taxi drivers in Greece?

Short Answer

Expert verified
The 95% confidence interval for the population mean following distance (μ) for taxi drivers in Greece while talking on a mobile phone is between 2.887 meters and 3.513 meters. To generalize this confidence interval to all taxi drivers in Greece, we must assume that the sample of 50 taxi drivers is representative of the entire population of taxi drivers in Greece, meaning the participants have been randomly selected and not biased in any way.

Step by step solution

01

Sample size, sample mean, and sample standard deviation

We know the sample size (n) is 50 taxi drivers, the sample mean (x̄) is 3.20 meters, and the sample standard deviation (s) is 1.11 meters. #Step 2: Calculate the margin of error#
02

Margin of error

To calculate the margin of error, we will use the formula: Margin of Error = \(t_{\frac{\alpha}{2}} \frac{s}{\sqrt{n}}\) For a 95% confidence interval, the critical value (t-score) can be found using a t-distribution table, or a calculator for the t-distribution. The degrees of freedom (df) are equal to n - 1, which is 49 in this case. \(t_{\frac{\alpha}{2}}\) = 2.010 Now we can compute the margin of error: Margin of Error = (2.010) \(\frac{1.11}{\sqrt{50}}\) ≈ 0.313 #Step 3: Construct the confidence interval#
03

Confidence interval

Now that we have the margin of error (0.313), we can construct the 95% confidence interval using the sample mean (3.20): Confidence Interval = (3.20 - 0.313, 3.20 + 0.313) = (2.887, 3.513) #Step 4: Interpret the confidence interval#
04

Interpretation

We are 95% confident that the true population mean following distance (μ) for taxi drivers in Greece while talking on a mobile phone is between 2.887 meters and 3.513 meters. #Step 5: Identify the assumption for generalization#
05

Assumption for generalization

In order to generalize this confidence interval to the population of all taxi drivers in Greece, we must assume that the sample of 50 taxi drivers is representative of the entire population of taxi drivers in Greece. That means, the participants must be randomly selected and not biased in any way.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Mean
The sample mean, often represented by the symbol \( \bar{x} \), is the average value of a sample—a smaller, selected set of data from a population. In our exercise involving Greek taxi drivers, the sample mean following distance was found to be 3.20 meters. Understanding the sample mean is crucial because it serves as an estimator of the population mean, \( \mu \), which we are often interested in determining.

However, it's important to bear in mind that the sample mean is just an estimate and not an exact measure of the population mean. To enhance its credibility, the sample should be as representative of the population as possible, and the sample size should be sufficiently large to reduce the variability of the mean.
Sample Standard Deviation
The sample standard deviation, denoted as \( s \), is a measure of the amount of variation or dispersion in a set of values. In our example, the sample standard deviation of the following distance was 1.11 meters. This figure tells us about the degree to which individual measurements of following distance deviate from the sample mean.

A smaller standard deviation indicates that the data points tend to be closer to the sample mean, whereas a larger standard deviation indicates that the data points are spread out over a wider range of values. Knowing the standard deviation assists us in understanding the uncertainty inherent in our sample statistics, which is essential when constructing confidence intervals.
T-Distribution
The t-distribution, also known as Student's t-distribution, comes into play when estimating population parameters, like the mean, from a small sample size with an unknown population standard deviation. The t-distribution is similar to the normal distribution, but with fatter tails, which gives more leeway for extreme values.

In the case of the taxi drivers, since the sample size is 50—which is considered relatively small—the t-distribution is an appropriate model for constructing the confidence interval. T-distributions are characterized by degrees of freedom (df), calculated as the sample size minus one (\( n - 1 \)). Therefore, with 50 drivers, our degrees of freedom are 49. The t-distribution allows us to account for the extra uncertainty of using a sample standard deviation instead of the true population standard deviation when estimating the margin of error for our confidence interval.
Margin of Error
The margin of error reflects the range around the sample mean within which we can expect to find the population mean a certain percentage of the time. It is linked to the confidence level, which represents how sure we are that the parameter lies within this range. A 95% confidence level, for example, suggests that if we were to take many samples and construct confidence intervals for each, we would expect about 95% of them to contain the true population mean.

In the example of the taxi drivers, we calculated a margin of error of approximately 0.313 meters. This means that we can be 95% confident that the actual population mean of the following distance while talking on a mobile phone lies within 0.313 meters of our sample mean on either side. When we give the confidence interval, we are essentially saying that we are 95% confident that the true population mean falls within that interval.

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Most popular questions from this chapter

The report "Majoring in Money: How American College Students Manage Their Finances" (SallieMae, \(2016,\) www.news/salliemae.com, retrieved December 24,2106\()\) includes data from a survey of college students. Each person in a representative sample of 793 college students was asked if they had one or more credit cards and if so, whether they paid their balance in full each month. There were 500 who paid in full each month. For this sample of 500 students, the sample mean credit card balance was reported to be \(\$ 825 .\) The sample standard deviation of the credit card balances for these 500 students was not reported, but for purposes of this exercises, suppose that it was \(\$ 200\). Is there convincing evidence that college students who pay their credit card balance in full each month have mean balance that is lower than \(\$ 906\), the value reported for all college students with credit cards? Carry out a hypothesis test using a significance level of 0.01 .

Explain the difference between \(\bar{x}\) and \(\mu_{\vec{x}}\).

Suppose that a random sample of 50 cans of a particular brand of fruit juice is selected, and the amount of juice (in ounces) in each of the cans is determined. Let \(\mu\) denote the mean amount of juice for the population of all cans of this brand. Suppose that this sample of 50 results in a \(95 \%\) confidence interval for \(\mu\) of (7.8,9.4) a. Would a \(90 \%\) confidence interval have been narrower or wider than the given interval? Explain your answer. b. Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 7.8 and 9.4 . Is this statement correct? Why or why not? c. Consider the following statement: If the process of selecting a random sample of size 50 and then calculating the corresponding \(95 \%\) confidence interval is repeated 100 times, exactly 95 of the resulting intervals will include \(\mu\). Is this statement correct? Why or why not?

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations. (Hint: See discussion on page \(594 .\) ) a. Upper-tailed test, \(\mathrm{df}=8, t=2.0\) b. Lower-tailed test, \(\mathrm{df}=10, t=-2.4\) c. Lower-tailed test, \(n=22, t=-4.2\) d. Two-tailed test, \(\mathrm{df}=15, t=-1.6\)

A manufacturing process is designed to produce bolts with a diameter of 0.5 inches. Once each day, a random sample of 36 bolts is selected and the bolt diameters are recorded. If the resulting sample mean is less than 0.49 inches or greater than 0.51 inches, the process is shut down for adjustment. The standard deviation of bolt diameters is 0.02 inches. What is the probability that the manufacturing line will be shut down unnecessarily? (Hint: Find the probability of observing an \(\bar{x}\) in the shutdown range when the actual process mean is 0.5 inches.)

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